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I want to know whether it is true that if a real sequence $\{x_n\}_{n=1}^\infty$ satisfies $\lim\limits_{n\to\infty} n|x_n-x_{n+1}|=0$ then it converges.

I guess it is false but I can't find a counterexample.

For your reference, this is from a deeper question ask for the range of $\alpha$ such that the following statement holds:

If a sequence $\{x_n\}_{n=1}^\infty$ of real numbers satisfies $\lim\limits_{n\to\infty}n^\alpha|x_n-x_{n+1}|=0$, then it is a convergent sequence.

Using the fact that $\sum_{k=1}^\infty k^{-\alpha}$ converges iff $\alpha>1$, I managed to find the necessary condition $\alpha\geq 1$ (by considering $x_n=\sum_{k=1}^{n-1}k^{-(\alpha+\epsilon)}$) and the sufficient condition $\alpha>1$ (by Cauchy's criterion). However, I stuck on the case $\alpha=1$, for which the same trick does not work. Therefore I am asking this question as a special case of the more general question I am solving.

Thanks for any answers.

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    $\begingroup$ Here's a counterexample: $\sum_{2}^\infty\frac{1}{n\log n}$, see math.stackexchange.com/questions/574503/… $\endgroup$ Commented 20 hours ago
  • $\begingroup$ @anonymous67 Thank you, it is indeed! Can't believe I have missed such an obvious one. Would you like to put you comment as an answer? $\endgroup$ Commented 20 hours ago
  • $\begingroup$ Sure. The case $\alpha=1$ always calls for the function $\log n$. $\endgroup$ Commented 20 hours ago
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    $\begingroup$ $\lim\limits_{n\to\infty} n|x_n-x_{n+1}|=0$ with the convergence of $$\lim_{n\to\infty} \frac{x_1+x_2+\cdots+x_n}{n}$$ can make $\{x_n\}$ convergent. $\endgroup$ Commented 20 hours ago
  • $\begingroup$ Properly, convergence follows from $n^{\color{blue}{1+h}}\to0$, where $h$ is a positive constant. But in the answer, we have $\ln n<<n^h$ for all positive $h$. $\endgroup$ Commented 2 hours ago

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Here's a counterexample: $\sum_{2}^\infty\frac{1}{n\log n}$, see this question.

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